HSC 2013 MX2 Marathon (archive) (2 Viewers)

seanieg89 · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Got up to iii, faced a problem:

$\langle z,w \rangle = t \langle w, w \rangle \\ \\ |\langle z,w \rangle|^2=t \langle w, z \rangle \sum_{j=1}^{n} |w_j|^2 \\ \\ \therefore \langle tw,z \rangle \ \ \in \mathbb{R} \\ \\ \therefore tw=z \\ \therefore $EQUALITY, not inequality for the expression we are trying to prove$$

Am I missing something here?

(Also I am very close to solving the last question)

Why does $\langle tw,z\rangle \in \mathbb{R}$ imply that $tw=z$ ?

Sy123 · Dec 9, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Why does $\langle tw,z\rangle \in \mathbb{R}$ imply that $tw=z$ ?

True that, well is this a good justification?

$|\langle z,w \rangle|^2=\langle tw, z \rangle \sum_{j=1}^n |w_j|^2$

$\langle tw, z \rangle \in \mathbb{R}$

$\langle z-tw, z-tw \rangle \geq 0$

$\therefore \langle tw, z \rangle \leq \langle z, z \rangle$

$\therefore |\langle z, w \rangle |^2 \leq \langle z, z \rangle \langle w, w \rangle$

$\therefore |\langle z, w \rangle| \leq \left(\sum_{j=1}^n|z_j|^2 \right)^{1/2}\left(\sum_{j=1}^n |w_j|^2 \right)^{1/2}$

Not sure if my reasoning is correct

Shadowdude · Dec 9, 2012

Re: HSC 2013 4U Marathon

cutemouse said:
Lol P Brown... Did you do it this year?

uh huh

Good times.

seanieg89 · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
True that, well is this a good justification?

$|\langle z,w \rangle|^2=\langle tw, z \rangle \sum_{j=1}^n |w_j|^2$

$\langle tw, z \rangle \in \mathbb{R}$

$\langle z-tw, z-tw \rangle \geq 0$

$\therefore \langle tw, z \rangle \leq \langle z, z \rangle$

$\therefore |\langle z, w \rangle |^2 \leq \langle z, z \rangle \langle w, w \rangle$

$\therefore |\langle z, w \rangle| \leq \left(\sum_{j=1}^n|z_j|^2 \right)^{1/2}\left(\sum_{j=1}^n |w_j|^2 \right)^{1/2}$

Not sure if my reasoning is correct

Why is < tw , z > smaller than < z , z >?

Seems correct if you can justify this step

.

Sy123 · Dec 9, 2012

Re: HSC 2013 4U Marathon

seanieg89 said:
Why is < tw , z > smaller than < z , z >?

Seems correct if you can justify this step .

I think i figured out how to justify it hehe (its pretty simple, not sure why I couldn't see it earlier)

$$Consider$ \ \ \ \langle z, z \rangle - \langle tw, z \rangle = z_1 \bar{z_1} - tw_1 \bar{z_1} + z_2 \bar{z_2}- tw_2 \bar{z_2}+...+z_n \bar{z_n}-tw_n \bar{z_n} \\ \\ = \bar{z_1}(z_1-tw_1)+...+\bar{z_n}(z_n-tw_n)= \langle z-tw, z-tw \rangle \geq 0$

$\therefore \langle z, z \rangle \geq \langle tw, z \rangle$

Now for the last question (Ill post a full solution once I (if I) answer it))

Examine · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
$$i) Show that $ z^5-1=(z-1)(z^2-2z\cos \frac{2\pi}{5}+1)(z^2-2z\cos \frac{4\pi}{5}+1) \ \ \ \ \ \fbox{2}$

$$ii) Hence form a quadratic equation with roots $ \ \ \ \cos \frac{2\pi}{5} \ \ \ $ AND $ \ \ \ \cos \frac{4\pi}{5} \ \ \ \ \ \ \fbox{3}$

Difficulty: Moderate

It would also be great if people could rate the difficulty of their own question as I have just done.

$\\ $We know that $ z^5-1 = \left ( z - 1 \right ) \left ( z^4 + z^3 + z^2 + z + 1 \right ) $ as well as $ z^5 - 1 = (z-1) \left ( z^2 - 2z \cos \frac{2 \pi}{5} + 1 \right ) \left ( z^2 - 2z \cos \frac{4 \pi}{5} + 1 \right ) \\\\ $Equating the coefficients of $ z $, we have:$ \\\\ -2 \cos \frac{2 \pi}{5} - 2 \cos \frac{4 \pi}{5} = 1 $ and therefore $ \cos \frac{2 \pi}{5} + \cos \frac{4 \pi}{5} = - \frac{1}{2} \\\\ $Similarly, equating the coefficients of $ z^2 $, we have $ \\\\ 1 + 1 + 4 \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} = 1 $ and therefore $ \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{5} = - \frac{1}{4} \\\\ $We now have the sum and product of roots of a quadratic with roots $ \cos \frac{2 \pi}{5} $ and $ \cos \frac{4 \pi}{5}. $ Thus, the quadratic equation is $ 4x^2 + 2x - 1 = 0.\\\\$

$\\ $ Solving this, we have $ x = \frac{-2 \pm \sqrt{4 - 4(4)(-1)}}{8} = \frac{-2 \pm \sqrt{20}}{8} = \frac{-1 \pm \sqrt{5}}{4} \\\\ $However, we know that $ \cos \frac{2 \pi}{5} > 0 $ since we are working in the first quadrant, and $ \cos \frac{4 \pi}{5} < 0 $ since we are working in the second quadrant. Therefore, we conclude that $ \cos \frac{2 \pi}{5} = \frac{-1 + \sqrt{5}}{2} $ and $ \cos \frac{4 \pi}{5} = \frac{-1 - \sqrt{5}}{2}$

Examine · Dec 9, 2012

Re: HSC 2013 4U Marathon

Shadowdude said:
Nice.

$\textup{Expand }(4+i)(5+3i)\textup{ and hence show that }\frac{\pi}{4} = \tan^{-1}\left ( \frac{1}{4} \right ) + \tan^{-1}\left (\frac{3}{5} \right )$

bit harder.

Challenge: Extend this question further so we have expressions like arctan(...) + arctan(...) = pi/4, where (...) are functions in terms of x.

RealiseNothing · Dec 9, 2012

Re: HSC 2013 4U Marathon

Examine said:
Challenge: Extend this question further so we have expressions like arctan(...) + arctan(...) = pi/4, where (...) are functions in terms of x.

Hi carrot.

Sy123 · Dec 9, 2012

Re: HSC 2013 4U Marathon

Examine said:
Challenge: Extend this question further so we have expressions like arctan(...) + arctan(...) = pi/4, where (...) are functions in terms of x.

Nice work on the above question, but for this, we can arrive at two variables:

$\tan^{-1} \frac{y}{x} + \tan^{-1} \frac{x-y}{x+y} = \frac{\pi}{4}$

Apply the tangent function to everything:

$y/x + \frac{x-y}{x+y} = 1- \frac{y(x-y)}{x+y}$

Simplify and we get:

$0=0$

Hence I think it can only be expressed as two variables not just in terms of single function x.

===========================

An easier one:

$$The polynomial equation $ S(x)=x^n - n x+1 $ has roots $ a_1, a_2, a_3,...,a_n$

$$Find $ a_1^n+a_2^n+a_3^n+...+a_n^n$

(I edited it, sorry for the inconvenience)

HeroicPandas · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
===========================

An easier one:

$$The polynomial equation $ S(x)=x^n - n x+1 $ has roots $ a_1, a_2, a_3,...,a_n$

$$Find $ a_1^n+a_2^n+a_3^n+...+a_n^n$

(I edited it, sorry for the inconvenience)

$a_1^n+a_2^n+a_3^n+...+a_n^n$ = n?

amen

Sy123 · Dec 9, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$a_1^n+a_2^n+a_3^n+...+a_n^n$ = n?

Close but not quite

HeroicPandas · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
Close but not quite

$a_1^n+a_2^n+a_3^n+...+a_n^n = -n?$

braintic · Dec 9, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$a_1^n+a_2^n+a_3^n+...+a_n^n$ = n?

amen

By substitution, (a_i)^n=n(a_i)-1
Summing: Sigma{(a_i)^n} = nSigma{(a_i)}-n

(If n>2) = 0-n=-n
(If n=2) = n(n)-n = 2

Sy123 · Dec 9, 2012

Re: HSC 2013 4U Marathon

HeroicPandas said:
$a_1^n+a_2^n+a_3^n+...+a_n^n = -n?$

Correct

seanieg89 · Dec 9, 2012

Re: HSC 2013 4U Marathon

Sy123 said:
I think i figured out how to justify it hehe (its pretty simple, not sure why I couldn't see it earlier)

$$Consider$ \ \ \ \langle z, z \rangle - \langle tw, z \rangle = z_1 \bar{z_1} - tw_1 \bar{z_1} + z_2 \bar{z_2}- tw_2 \bar{z_2}+...+z_n \bar{z_n}-tw_n \bar{z_n} \\ \\ = \bar{z_1}(z_1-tw_1)+...+\bar{z_n}(z_n-tw_n)= \langle z-tw, z-tw \rangle \geq 0$

$\therefore \langle z, z \rangle \geq \langle tw, z \rangle$

Now for the last question (Ill post a full solution once I (if I) answer it))

You are close to the right idea, the key is that < z-tw, z-tw > is non-negative. Expand this expression out and you will get what you want

.

The final part of the question should be easier once you interpret it correctly.

Synchronised · Dec 11, 2012

Re: HSC 2013 4U Marathon

$\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx$

SpiralFlex · Dec 11, 2012

Re: HSC 2013 4U Marathon

Synchronised said:
$\int_{0}^{1}\frac{ln(1+x)}{1+x^{2}}dx$

Don't be mean....4U questions only here. No Polylogarithms please.

Synchronised · Dec 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Don't be mean....4U questions only here. No Polylogarithms please.

It is a 4U question.

cutemouse · Dec 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Don't be mean....4U questions only here. No Polylogarithms please.

What's a "Polylogarithm"?

And that question can be done using 4U methods.

seanieg89 · Dec 11, 2012

Re: HSC 2013 4U Marathon

SpiralFlex said:
Don't be mean....4U questions only here. No Polylogarithms please.

Evaluating the definite integral is well within the scope of MX2, the substitution

$x=\frac{1-u}{1+u}$

kills it.

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